From the series of 200 lines beginning alternately with “3” then “1” from the previous post, I removed all lines that began with a “1”. That left 100 lines beginning with a “3” followed by one or more “4”s. I removed from each line the initial “3”.
What remains is a series of 100 lines with one or more “4”s.
Do you see patterns? Can you figure out a rule?
I don’t know the answer: hypothetically, TM #4 (alias “Chaotic”) of Marxen and Buntrock might be computing a sequence of integers….
4
44
444
4444
44444
444444
4444444
44444444
444444444
44
44444444444
444444444444
4444444444444
44444444444444
444444444444444
4444444444444444
4
44
4444444444444444444
4444
444444444444444444444
444444
44444444444444444444444
444444444444444444444444
4444444444444444444444444
44
444444444444444444444444444
4444444444444444444444444444
44444444444444444444444444444
444444444444444444444444444444
4444444444444444444444444444444
44444444444444444444444444444444
4
44
444
4444
4444444444444444444444444444444444444
444444
4444444
4444444444444444444444444444444444444444
444444444
44
4444444444444444444444444444444444444444444
444444444444
444444444444444444444444444444444444444444444
44444444444444
44444444444444444444444444444444444444444444444
ETC.
===========================================
Added: Friday September 1st, 2017 at 10am
Below, for every line number from 1 to 100, we give the number
of 4s on that line:
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 2
11 11
12 12
13 13
14 14
15 15
16 16
17 1
18 2
19 19
20 4
21 21
22 6
23 23
24 24
25 25
26 2
27 27
28 28
29 29
30 30
31 31
32 32
33 1
34 2
35 3
36 4
37 37
38 6
39 7
40 40
41 9
42 2
43 43
44 12
45 45
46 14
47 47
48 48
49 1
50 2
51 51
52 4
53 53
54 6
55 55
56 56
57 57
58 2
59 59
60 60
61 61
62 62
63 63
64 64
65 1
66 2
67 3
68 4
69 5
70 6
71 7
72 72
73 9
74 2
75 11
76 12
77 77
78 14
79 15
80 80
81 1
82 2
83 19
84 4
85 85
86 6
87 23
88 88
89 25
90 2
91 91
92 28
93 93
94 30
95 95
96 96
97 1
98 2
99 3
100 4
===========================================
Added: Friday September 1st, 2017 at 1 pm
Below, for every line number from 1 to 100 ( = A), we give the number
of 4s on that line ( = B) , and the difference
C = A – B.
C is a multiple of 8…
A B C
————-
1 1 0
2 2 0
3 3 0
4 4 0
5 5 0
6 6 0
7 7 0
8 8 0
9 9 0
10 2 8
11 11 0
12 12 0
13 13 0
14 14 0
15 15 0
16 16 0
17 1 16
18 2 16
19 19 0
20 4 16
21 21 0
22 6 16
23 23 0
24 24 0
25 25 0
26 2 24
27 27 0
28 28 0
29 29 0
30 30 0
31 31 0
32 32 0
33 1 32
34 2 32
35 3 32
36 4 32
37 37 0
38 6 32
39 7 32
40 40 0
41 9 32
42 2 40
43 43 0
44 12 32
45 45 0
46 14 32
47 47 0
48 48 0
49 1 48
50 2 48
51 51 0
52 4 48
53 53 0
54 6 48
55 55 0
56 56 0
57 57 0
58 2 56
59 59 0
60 60 0
61 61 0
62 62 0
63 63 0
64 64 0
65 1 64
66 2 64
67 3 64
68 4 64
69 5 64
70 6 64
71 7 64
72 72 0
73 9 64
74 2 72
75 11 64
76 12 64
77 77 0
78 14 64
79 15 64
80 80 0
81 1 80
82 2 80
83 19 64
84 4 80
85 85 0
86 6 80
87 23 64
88 88 0
89 25 64
90 2 88
91 91 0
92 28 64
93 93 0
94 30 64
95 95 0
96 96 0
97 1 96
98 2 96
99 3 96
100 4 96
=========================================================
I arranged the 100 values of C := A – B in rows of 8:
0 0 0 0 0 0 0 0
0 8 0 0 0 0 0 0
16 16 0 16 0 16 0 0
0 24 0 0 0 0 0 0
32 32 32 32 0 32 32 0
32 40 0 32 0 32 0 0
48 48 0 48 0 48 0 0
0 56 0 0 0 0 0 0
64 64 64 64 64 64 64 0
64 72 64 64 0 64 64 0
80 80 64 80 0 80 64 0
64 88 0 64 0 64 0 0
96 96 96 96
==================================
16 8 32 16 64 16 32 – GCD
==================================
2 1 4 2 8 2 4 – GCD/8
==================================
=======================================================
I extended the previous grouping in rows of 8 to the
first 128 lines, as opposed to the first 100 lines of
the hypothetical sequence:
0 0 0 0 0 0 0 0
0 8 0 0 0 0 0 0
16 16 0 16 0 16 0 0
0 24 0 0 0 0 0 0
32 32 32 32 0 32 32 0
32 40 0 32 0 32 0 0
48 48 0 48 0 48 0 0
0 56 0 0 0 0 0 0
64 64 64 64 64 64 64 0
64 72 64 64 0 64 64 0
80 80 64 80 0 80 64 0
64 88 0 64 0 64 0 0
96 96 96 96 0 96 96 0
96 104 0 96 0 96 0 0
112 112 0 112 0 112 0 0
0 120 0 0 0 0 0 0
================================================
I’m adding a table in graphics/picture format of the
16 rows of 8, as I don’t know how to get fixed-width fonts
in WordPress:
=====================================
Added Tuesday September 5, 2017 at 5:50 pm
We now look at these numbers in base 8.
Because they are all multiples of 8, we first divide by 8.
Secondly, in the newgroup sci.math, I observed that the terms
modulo 64 have a periodicity of 64, and the terms modulo 512
have a periodicity of 512, up to and including the 700th term at least.
Therefore, to further verify the (terms mod 64) period of 64, we can look only
at the least significant 2 base 8 digits, and as the last base 8 digit is always 0, we need only tabulate the second to last base 8 digit for what turns out to be
the first 704 terms. Period 64 checks out all right:
Below, we have the 2nd base 8 (octal) digit
from the left. One can check that there’s
a period of length 64 in these base 8 digits.
We show 88 rows of length 8 for a total
of 704 terms.
0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 2 0 2 0 2 0 0
0 3 0 0 0 0 0 0
4 4 4 4 0 4 4 0
4 5 0 4 0 4 0 0
6 6 0 6 0 6 0 0
0 7 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 2 0 2 0 2 0 0
0 3 0 0 0 0 0 0
4 4 4 4 0 4 4 0
4 5 0 4 0 4 0 0
6 6 0 6 0 6 0 0
0 7 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 2 0 2 0 2 0 0
0 3 0 0 0 0 0 0
4 4 4 4 0 4 4 0
4 5 0 4 0 4 0 0
6 6 0 6 0 6 0 0
0 7 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 2 0 2 0 2 0 0
0 3 0 0 0 0 0 0
4 4 4 4 0 4 4 0
4 5 0 4 0 4 0 0
6 6 0 6 0 6 0 0
0 7 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 2 0 2 0 2 0 0
0 3 0 0 0 0 0 0
4 4 4 4 0 4 4 0
4 5 0 4 0 4 0 0
6 6 0 6 0 6 0 0
0 7 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 2 0 2 0 2 0 0
0 3 0 0 0 0 0 0
4 4 4 4 0 4 4 0
4 5 0 4 0 4 0 0
6 6 0 6 0 6 0 0
0 7 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 2 0 2 0 2 0 0
0 3 0 0 0 0 0 0
4 4 4 4 0 4 4 0
4 5 0 4 0 4 0 0
6 6 0 6 0 6 0 0
0 7 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 2 0 2 0 2 0 0
0 3 0 0 0 0 0 0
4 4 4 4 0 4 4 0
4 5 0 4 0 4 0 0
6 6 0 6 0 6 0 0
0 7 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 2 0 2 0 2 0 0
0 3 0 0 0 0 0 0
4 4 4 4 0 4 4 0
4 5 0 4 0 4 0 0
6 6 0 6 0 6 0 0
0 7 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 2 0 2 0 2 0 0
0 3 0 0 0 0 0 0
4 4 4 4 0 4 4 0
4 5 0 4 0 4 0 0
6 6 0 6 0 6 0 0
0 7 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 2 0 2 0 2 0 0
0 3 0 0 0 0 0 0
4 4 4 4 0 4 4 0
4 5 0 4 0 4 0 0
6 6 0 6 0 6 0 0
0 7 0 0 0 0 0 0
In a picture:
